# Fatou's Lemma and Almost Sure Convergence (Pt. 2)

This is an extension to the question I asked earlier here. The overarching question was the following homework problem,

Let $X_{n} \rightarrow X$ in probability. Show that $\liminf_{n} \mathbb{E}[X_{n}] \ge \mathbb{E}[X]$.

Obviously, this is very similar to the statement of Fatou's Lemma. I also know the following,

$X_{n} \rightarrow X$ in probability if and only if for all subsequences $X_{n(m)}$ of $X_{n}$ there exists a sub-subsequence $X_{n(m_{k})} \rightarrow X$ almost surely.

Given what was answered in my last question, it is trivial to show that Fatou's lemma applies to all of these sub-subsequences $X_{n(m_{k})}$, but how can I bring this back to $X_{n}$ itself, being that it lacks almost sure convergence to $X$?

EDIT Another condition is that $X_n \ge 0$

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Missing hypotheses? As stated, this is false. For a counterexample, assume that $P(X_n=0)=1-p_n$ and $P(X_n=x_n)=p_n$ for given $x_n$ and $p_n$. Then $X_n\to X=0$ in probability as soon as $p_n\to0$, but $E(X)=0$ and $E(X_n)=x_np_n$ can be anything one wants. – Did Sep 13 '11 at 19:00
You're right, I forgot to add that all Random Variables must be non negative. – duckworthd Sep 14 '11 at 2:59

## 1 Answer

Edit: My first version was a bit too complicated, so here's a better version:

If $C = \liminf\limits_{n\to\infty}\;{\mathbb{E}[X_n]} = \infty$ there's nothing to prove, so assume that $C \lt \infty$.

We may choose a subsequence $X_{n(m)}$ such that $\mathbb{E}[X_{n(m)}] \to C = \liminf\limits_{n\to\infty}\;{\mathbb{E}[X_{n}]}$ by the definition of the $\liminf$. As you stated in your question, there's a sub-subsequence $X_{n(m_k)}$ such that $X_{n(m_k)} \to X$ a.e. since $X_n \to X$ in measure. As $X_n \geq 0$ a.e., the pointwise a.e. Fatou lemma gives us $\mathbb{E}[X] \leq \liminf\limits_{k \to \infty}\;E[X_{n(m_k)}]= \liminf\limits_{n\to\infty}\;{\mathbb{E}[X_n]}$, as desired.

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What do you mean by $\mathbb{E}[X_{n(m)}] < C + \epsilon$? Should that be $\lim_{m \rightarrow \infty} \mathbb{E}[X_{n(m)}] < C + \epsilon$? I think the proof goes through if that was your intent. – duckworthd Sep 14 '11 at 17:58
I meant what I wrote. Note that $\mathbb{E}[X_{n}]$ is a sequence of real numbers that accumulates to $C$, thus for every $\varepsilon \gt 0$ there must be infinitely many $X_n$'s such that $\mathbb{E}[X_n]$ is smaller than $C + \varepsilon$. Just pick your subsequence among those. But you can of course also do what you suggest. – t.b. Sep 14 '11 at 18:10
@duckworthd: I've updated my answer following your suggestion. The previous version was correct but a bit too complicated, sorry about that. I hope that it's fine now. – t.b. Sep 15 '11 at 2:25